### ... by: Nono

Question like Str8tsFan.

the last solver version 1.95 does not eliminate the 8 in J1.

No problem for good sudoku players !

Andrew Stuart writes:

Quite correct. There was some very odd code in there that stopped J1 from being eliminated because the link had already eliminated in the box. Can't say why, must have had a brain fart that day. Diagram and solver updated.

### ... by: Str8tsFan

I hope Andrew will ever read (and answer) these comments... About the second example:

(1) Using the link to the solver leads to a sudoku with a tiny little difference: B9 has an additional candidate 6, which is missing at the example. At the solver this candidate will be eliminated with the very same example:

"- Off-chain candidate 6 taken off B9 - weak link: B5 to B8"

(2) What about the candidate 8 at J1? As far as I can understand the theory, the weak link "D1 to F1" should not only eliminate the 8s at F2 and F3 (weak link in same box) but also the 8 at J1 (weak link in same column). More interesting: the solver doesn't eliminate that 8 at J1 either. Why? Did I make any mistake, or is it a flaw of the solver? As far as I understood, a weak link can be part of two entities (box and row or box and column) and thus should be able to eliminate candidates at both entities, maybe the solver fails to check the second entity?

Andrew Stuart writes:

Thanks for the detailed hints. Both points correct. I've updated the diagram and solver.

### ... by: Mr Turner

The last paragraph seems flawed since there is an 8 at F8. The 8s at F2 and F3 could be colored off as an inference from D1 being on. This implies F8 is on. F8 is also inferred to be on from D8 being off.
Andrew Stuart writes:

Correct. I've removed that paragraph now. Doh.

### ... by: Mr Archibald

I call this the 6 pack.

the first three in a line on rows 1,2 is the some as the last six on row 3. this can work upsidedown and backtofront and sideways.

a b c d e f g h i

i h g f e d c b a

d e f a b c i h g